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a) \(5^6:5^5+\left(\dfrac{4}{9}\right)^0=5^{6-5}+1=5+1=6\)
b) \(\left(\dfrac{3}{7}\right)^{21}:\left(1-\dfrac{40}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^6\)
\(=\left(\dfrac{3}{7}\right)^{21-6}=\left(\dfrac{3}{7}\right)^{15}\)
c) \(\left(\dfrac{2}{3}\right)^3-\left(\dfrac{-52}{3}\right)^0+\dfrac{4}{9}\)
\(=\dfrac{8}{27}-1+\dfrac{4}{9}\)
\(=\dfrac{8-27+12}{27}=-\dfrac{7}{27}\)
\(a)5^6:5^5+\left(\dfrac{4}{9}\right)^0=5^1+1=6\)
\(b,\left(\dfrac{3}{7}\right)^{21}:\left(1-\dfrac{40}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{49-40}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^3=\left(\dfrac{3}{7}\right)^{21}:[\left(\dfrac{3}{7}\right)^2]^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^6=\left(\dfrac{3}{7}\right)^{21-6}\)
\(=\left(\dfrac{3}{7}\right)^{15}\)
\(c,3.\left(\dfrac{2}{3}\right)^3-\left(\dfrac{-52}{3}\right)^0+\dfrac{4}{9}\)
\(=3.\dfrac{8}{27}-1+\dfrac{4}{9}\)
\(=\dfrac{8}{9}-1+\dfrac{4}{9}\)
\(=\dfrac{8-9+4}{9}=\dfrac{1}{3}\)
a) Ta có: \(\left(\dfrac{9}{25}-2\cdot18\right):\left(3\dfrac{4}{5}+0.2\right)\)
\(=\left(\dfrac{9}{25}-36\right):\left(\dfrac{19}{5}+\dfrac{1}{5}\right)\)
\(=\left(\dfrac{9}{25}-\dfrac{900}{25}\right):\dfrac{20}{5}\)
\(=\dfrac{-891}{25}\cdot\dfrac{1}{4}\)
\(=-\dfrac{891}{100}\)
b) Ta có: \(\dfrac{3}{8}\cdot19\dfrac{1}{3}+\dfrac{3}{8}\cdot33\dfrac{1}{3}\)
\(=\dfrac{3}{8}\cdot\dfrac{58}{3}+\dfrac{3}{8}\cdot\dfrac{100}{3}\)
\(=\dfrac{58}{8}+\dfrac{100}{8}\)
\(=\dfrac{158}{8}=\dfrac{79}{4}\)
c) Ta có: \(15\cdot\left(-\dfrac{2}{3}\right)^2-\dfrac{7}{3}\)
\(=15\cdot\dfrac{4}{9}-\dfrac{7}{3}\)
\(=\dfrac{20}{3}-\dfrac{7}{3}\)
\(=\dfrac{13}{3}\)
d) Ta có: \(\dfrac{1}{2}\sqrt{64}-\sqrt{\dfrac{4}{25}}+\left(-1\right)^{2007}\)
\(=\dfrac{1}{2}\cdot8-\dfrac{2}{5}-1\)
\(=4-1-\dfrac{2}{5}\)
\(=3-\dfrac{2}{5}\)
\(=\dfrac{15}{5}-\dfrac{2}{5}=\dfrac{13}{5}\)
e) Ta có: \(\left(-\dfrac{5}{2}\right)^2:\left(-15\right)-\left(0.45+\dfrac{3}{4}\right)\cdot\left(-1\dfrac{5}{9}\right)\)
\(=\dfrac{25}{4}\cdot\dfrac{-1}{15}-\left(\dfrac{9}{20}+\dfrac{15}{20}\right)\cdot\dfrac{-14}{9}\)
\(=\dfrac{-25}{60}-\dfrac{24}{20}\cdot\dfrac{-14}{9}\)
\(=\dfrac{-25}{60}+\dfrac{28}{15}\)
\(=\dfrac{-25}{60}+\dfrac{112}{60}\)
\(=\dfrac{87}{60}=\dfrac{29}{20}\)
f) Ta có: \(\left(-\dfrac{1}{3}\right)-\left(-\dfrac{3}{5}\right)^0+\left(1-\dfrac{1}{2}\right)^2:2\)
\(=-\dfrac{1}{3}-1+\left(\dfrac{1}{2}\right)^2\cdot\dfrac{1}{2}\)
\(=\dfrac{-4}{3}+\dfrac{1}{4}\cdot\dfrac{1}{2}\)
\(=\dfrac{-4}{3}+\dfrac{1}{8}\)
\(=\dfrac{-32}{24}+\dfrac{3}{24}=\dfrac{-29}{24}\)
g) Ta có: \(\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{4}\right)^{20}\)
\(=\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{2}\right)^{40}\)
\(=\left(\dfrac{1}{2}\right)^{55}\)
\(=\dfrac{1}{2^{55}}\)
h) Ta có: \(\dfrac{5^4\cdot20}{25^5\cdot4^5}\)
\(=\dfrac{5^4\cdot5\cdot2^2}{5^{10}\cdot2^{10}}\)
\(=\dfrac{5^5}{5^{10}}\cdot\dfrac{2^2}{2^{10}}\)
\(=\dfrac{1}{5^5}\cdot\dfrac{1}{2^8}\)
\(=\dfrac{1}{800000}\)
`@` `\text {Ans}`
`\downarrow`
`a)`
`-9/34*17/4`
`=`\(\dfrac{-9}{17\cdot2}\cdot\dfrac{17}{4}\)
`=`\(-\dfrac{9}{2}\cdot\dfrac{1}{4}\)
`=`\(-\dfrac{9}{8}\)
`b)`
\(\dfrac{17}{15}\div\dfrac{4}{3}\)
`=`\(\dfrac{17}{15}\cdot\dfrac{3}{4}\)
`=`\(\dfrac{17}{3\cdot5}\cdot\dfrac{3}{4}\)
`=`\(\dfrac{17}{5}\cdot\dfrac{1}{4}\)
`=`\(\dfrac{17}{20}\)
`c)`
\(4\dfrac{1}{5}\div\left(-2\dfrac{4}{5}\right)\)
`=`\(4\dfrac{1}{5}\cdot\left(-\dfrac{5}{14}\right)\)
`=`\(\dfrac{21}{5}\cdot\left(-\dfrac{5}{14}\right)\)
`=`\(-\dfrac{21}{14}=-\dfrac{3}{2}\)
a) \(\dfrac{-9}{34}\cdot\dfrac{17}{4}\)
\(=\dfrac{-9\cdot17}{34\cdot4}\)
\(=-\dfrac{153}{136}\)
\(=\dfrac{9}{8}\)
b) \(\dfrac{17}{15}:\dfrac{4}{3}\)
\(=\dfrac{17}{15}\cdot\dfrac{3}{4}\)
\(=\dfrac{17\cdot3}{15\cdot4}\)
\(=\dfrac{51}{60}=\dfrac{17}{20}\)
c) \(4\dfrac{1}{5}:\left(-2\dfrac{4}{5}\right)\)
\(=\dfrac{21}{5}:-\dfrac{14}{5}\)
\(=\dfrac{21}{5}\cdot-\dfrac{5}{14}\)
\(=\dfrac{21\cdot-5}{5\cdot14}\)
\(=-\dfrac{105}{70}=\dfrac{3}{2}\)
\(E=\dfrac{98:\left(\dfrac{4}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\left(\dfrac{27}{25}-\dfrac{2}{25}\right)\cdot\dfrac{7}{4}}{\left(\dfrac{59}{9}-\dfrac{13}{4}\right)\cdot\dfrac{36}{17}}\\ E=\dfrac{98}{\dfrac{3}{5}}+\dfrac{\dfrac{7}{4}}{\dfrac{119}{36}\cdot\dfrac{36}{17}}\\ E=\dfrac{490}{3}+\dfrac{\dfrac{7}{4}}{7}=\dfrac{490}{3}+\dfrac{1}{4}=\dfrac{1963}{12}\)
bạn ơi chỗ kia mik nhìn hơi loạn tí bạn giải thích giúp mik với
f, \(\dfrac{2^9.4^{10}}{8^8}=\dfrac{2^9.\left(2^2\right)^{10}}{\left(2^3\right)^8}=\dfrac{2^9.2^{20}}{2^{24}}=\dfrac{2^{29}}{2^{24}}=2^5=32\)
a: \(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\dfrac{14}{25}+\dfrac{11}{25}+\dfrac{2}{7}=\dfrac{2}{7}\)
b: \(=\dfrac{3}{7}-\dfrac{5}{2}-\dfrac{3}{5}+\dfrac{4}{7}+\dfrac{3}{2}-\dfrac{2}{5}=1-1-1=-1\)
c: \(=\dfrac{4}{25}+\dfrac{7}{5}\cdot\dfrac{5}{2}-2=\dfrac{4}{25}+\dfrac{7}{2}-2=\dfrac{83}{50}\)
\(1,\\ a,2< 3\Rightarrow2^{30}< 3^{30}\Rightarrow-2^{30}>-3^{30}\\ b,6^{10}=6^{2\cdot5}=\left(6^2\right)^5=36^5>35^5\left(36>35\right)\)
\(2,\\ a,\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\dfrac{3^{10}\cdot5^5\cdot3^5}{5^6\cdot3^{14}}=\dfrac{3}{5}\\ b,\left(8x-1\right)^{2x+1}=5^{2x+1}\\ \Leftrightarrow8x-1=5\\ \Leftrightarrow x=\dfrac{3}{4}\)
Bài 2:
a: Ta có: \(\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}\)
\(=\dfrac{-3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^{14}}\)
\(=-\dfrac{3}{5}\)
b: Ta có: \(\left(8x-1\right)^{2x+1}=5^{2x+1}\)
\(\Leftrightarrow8x-1=5\)
\(\Leftrightarrow8x=6\)
hay \(x=\dfrac{3}{4}\)
1)(-1/2)^2:1/4-2.(-1/2)^3+căn 4
=1/4:1/4-2.-1/8+2
= 1-(-1/4)+2
=1+1/4+2=13/4
2) 3-(-6/7)^0+căn 9 :2
= 3-1+3:2
=3-1+3/2=7/2
3) (-2)^3+1/2:1/8-căn 25 + |-64|
= -8+4-5+64= 55
4) (-1/2)^4+|-2/3|-2007^0
= 1/16+2/3-1
= -13/48
5) = 178/495:623/495-17/60:119/120
= 2/7-2/7=0
6) [2^3.(-1/2)^3+1/2]+[25/22+6/25-3/22+19/25+1/2]
= [-1+1/2]+[(25/22-3/22)+(6/25+19/25)+1/2]
= -1/2+[1+1+1/2]
= -1/2+5/2=2
Mấy cái dấu chấm đó là nhân nha bn!
bài 1)
a) \(\dfrac{\left(-3\right)^{10}.15^5}{25^3.\left(-9\right)^7}\)
\(=\dfrac{\left(-3\right)^{10}.\left(3.5\right)^5}{\left(5^2\right)^3.\left(-3.3\right)^7}\)
\(=\dfrac{\left(-3\right)^{10}.3^5.5^5}{5^6.\left(-3\right)^7.3^7}\)
\(=\dfrac{\left(-3\right)^3.1.1}{5.1.3^2}\)
\(=\dfrac{-27.1.1}{5.1.9}\)
\(=\dfrac{-27}{45}\)
\(=\dfrac{-9}{15}\)
b)\(2^3+3.\left(\dfrac{1}{9}\right)^0-2^{-2}.4\left[\left(-2\right)^2:\dfrac{1}{2}\right].8\)
\(=8+3.1-\dfrac{1}{2^2}.4+\left[\left(4:\dfrac{1}{2}\right)\right].8\)
\(=8+3.1-\dfrac{1}{4}.4+\left[4.\dfrac{2}{1}\right].8\)
\(=8+3.1-\dfrac{1}{4}.4+8.8\)
\(=8+3-1+64\)
\(=11-1+64\)
\(=10+64\)
\(=74\)
Mình cảm ơn bạn nha .